1. Day 15 AGENDA:
QUIZ # minutes
Work Day PW Lesson #11:CLT
Begin Unit 1 Lesson #12
UNIT 1 TEST 1 Thurs; Rev on blog

2. Lesson #12: Confidence Intervals
Accel Precalc
Unit #1: Data Analysis
Lesson #12: Confidence Intervals
For Means
EQ: How do you use confidence intervals to estimate population means and proportions?

3. Terms to Recall
Statistics ---
measures of a sample; also called point estimates
Parameters
--- measures of a population

4. What are some statistics and parameters we have discussed thus far?
Measure Statistic Parameters 
Mean s 
Standard Deviation
In theory you will never know parameters. That is why you use statistics to estimate.

5. New Terms: Interval Estimates ---
interval of values used to estimate a parameter
Ex. average age of students might be in an interval
17.9 <  < 18.7 or
18.3 ± 0.4 years
0.4 is called the margin of error

6. Confidence Level
--- probability that the confidence interval will contain the true parameter.
specific interval estimate for a population parameter determined by using statistics obtained from a sample.
Confidence Interval ---

9. Handout Table C

10. See Table C for critical values t*= 1.645 90%, 95%, 99%
Use the Table C to
determine the
corresponding t* value
for each interval
when n is very large.
Confidence Level = 90%
t*= 1.645 5%
90% 5%
1.645
Common Confidence Intervals:
90%, 95%, 99%

11. Confidence Level = 95% t*= 1.96 Confidence Level = 99% t*= 2.576 2.5%
0.5%
0.5%
2.576

12. 0.95 P(z < 1.645) = Verify by finding the probability under the
curve at the given z values.
P(z < 1.645) =
0.95
RECALL: How do you find area when given a z-score?

13. P(z < 1.96) = 0.975 Verify by finding the probability under the
curve at the given z values.
P(z < 1.96) =
0.975
RECALL: How do you find area when given a z-score?

14. P(z < 2.575) = 0.995 Verify by finding the probability under the
curve at the given z values.
P(z < 2.575) =
0.995
RECALL: How do you find area when given a z-score?

15. AGENDA: Finish Unit 1 Lesson #12 Review for Test
UNIT 1 TEST 1 Fri; Rev on blog

16. Formula for Confidence Intervals For Means:
RECALL:
Standard Error

17. Uses a t-distribution and degrees of freedom where df = n - 1
Recall:
Central Limit Theorem
When the sample size is large enough,  95% of the sample means will fall within 1.96 standard deviations of the true parameter.

18. This means for a 95% confidence interval:

19. Write the general formula for:
Margin of Error

20. Write the general formula for:
Margin of Error

21. Ex 1 Find the t* values for the following then write the general formula for the confidence interval.
90% conf int when sample size is 22
Confidence Level = 90%
n = 22
df = 21
Now you must pay attention to sample size.

22. CL = 90%
n = 22
df = 21
t*= 1.721
Margin of Error

23. b) 95% conf int when sample size is 15
Confidence Level = 95%
n = 15
df = 14
t*= 2.145
Margin of Error

24. c) 99th percentile when sample size is 30
Confidence Level = 98% WHY?
n = 30
t*= 2.462
df = 29
Margin of Error

25. Confidence Interval Statement:
We are _______% confident the true
population mean __________________
___________________________ is
between _______ and _________ for
a sample size n = ______.

26. n = 50 df = 49 Must read t* at 40 t*= 2.021
In Class Examples. Place formula and work on your own paper. Make a conclusion statement in context of the problem.
A researcher wishes to estimate the average of money a person spends on lottery tickets each month. A sample of 50 people who play the lottery found the mean to $19 and the standard deviation to be Find the best point estimate of the population and the 95% confidence interval of the population mean.
Confidence Level = 95%
n = 50
df = 49 Must read
t* at 40
t*= 2.021
What is the margin of error?
± 1.944

27. of money a person spends on lottery tickets each month
Confidence Interval Statement: 95
We are _______% confident the true
population mean __________________
___________________________ is
between _______ and _________ for
a sample size n = ______.
of money a person spends
on lottery tickets each month
$17.06
$20.94
50 people

28. 2. The following data represent a sample of the assets (in millions of dollars) of 30 credit unions in southwestern Pennsylvania. Find the 90% confidence interval of the mean.
How would you find these
values if I didn’t give them to you?
CL = 90%
n = 30
df = 29
t*= 1.699
What is the margin of error?
± $4.468 million

29. assets of credit unions in southwestern Pennsylvania
Confidence Interval Statement: 90
We are _______% confident the true
population mean __________________
___________________________ is
between _______ and _________ for
a sample size n = ______.
assets of credit unions in southwestern Pennsylvania
$6.623 million
$15.56 million
30 credit unions

30. n = 30 df = 29 t*= 1.699 Confidence Level = 90%
3. A study of 30 marathon runners showed that they could run at an average of 7.8 miles per hour. The sample standard deviation is Find the point estimate for the mean of all runners. Based on the results, what minimum speed should a runner obtain to qualify in a marathon at a 90% confidence level.
Confidence Level = 90%
n = 30
df = 29
t*= 1.699
What is the margin of error?
± mph
Point Estimate
At a 90% confidence level and using a sample size of n= 30 runners, the minimum speed necessary for a runner to qualify for a marathon is 7.61 mph.

31. n = 10 df = 9 t*= 2.262 Confidence Level = 95%
4. Ten randomly selected automobiles were stopped and the tread depth of the right front tire was measured. The mean was 0.32 in and the standard deviation was 0.08 in. Find the 95% confidence interval of the mean depth. Assume that the variable is approximately normally distributed.
Confidence Level = 95%
n = 10
df = 9
t*= 2.262
What is the margin of error?
± in

32. 95 10 automobiles tread depth of the right front
Confidence Interval Statement: 95
We are _______% confident the true
population mean __________________
___________________________ is
between _______ and _________ for
a sample size n = ______.
tread depth of the right front
tire of a selected make of automobile
0.38 in
0.26 in
10 automobiles

33. n = 7 df = 6 t*= 3.707 Confidence Level = 99%
5. The data below represent a sample of the number of home fires started by candles for the past seven years. Find the 99% confidence interval for the mean number of home fires started by candles each year. Assume the variable is normally distributed.  
How do you find these
values if I didn’t give them to you?
Confidence Level = 99%
n = 7
df = 6
t*= 3.707
What is the margin of error?
± fires

34. number of home fires started by candles
Confidence Interval Statement: 99
We are _______% confident the true
population mean __________________
___________________________ is
between _______ and _________ for
a sample size n = ______.
number of home fires started by candles
4,785.2 fires
9,297.6 fires
7 years

35. Assignment: Practice Worksheet Confidence Intervals